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Let $R = \{(x, xe^n) : x \in \Bbb R,n \in \Bbb Z\}$. Prove that $R$ is an equivalence relation on the set of real numbers.

Edit: Sorry this is my first time. I am aware that in order to prove that it is an equivalence relation, I need to prove that it is reflexive, symmetric an transitive.

For reflexive I am unsure if this is correct but I have put:

$(x,x) \in R$ so, $x = xe^n$ so if we let $n = 0$ then, $x = x$

For symmetric, since $(x,y) ∈ R \implies (y,x) \in R$

$(x,xe^n) \implies (xe^n,x)$, so $y =xe^n$ , so $x = y/e^n$

for transitive, so far I wrote:

$(x,y) , (y,z) \implies (x,z)$

$(x,xe^n)$ and $(xe^n, e^n(xe^n)) \implies (x,e^n(xe^n))$ which can be simplified to, $( x,xe^{2n} )$ however I do not know where to go from here.

jnljjkbj
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    Please edit to include your efforts. – lulu Feb 15 '23 at 23:44
  • What did you try so far? – MathematicianByMistake Feb 15 '23 at 23:44
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    Hi! To avoid down-votes and close-votes, please provide us some context for this question, such as: (a) Is this homework? (b) If so, what course are you taking? (c) What specific topic are you covering at the moment? (d) What do you know that you think might be connected? (e) If you're stuck, what are you stuck on? For example, do you know what to apply, but don't know how to apply it, or do you not know what to apply? Please put these facts in your original post, not as responses to this comment, as comments may be deleted without warning. – Brian Tung Feb 15 '23 at 23:46
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    Hello. On top of what has already been suggested, making a little effort on typing the appropriate symbols will really help people help you. For instance, does x belong to the set of real numbers or is it any other set you are calling R just so that we are confused by the fact that it is the same name you've given your equivalence relation? – juan zaragoza Feb 15 '23 at 23:51
  • Good work proving reflexivity. The same approach proves the other properties as well. Try Symmetry next. If $x\sim y$ then we know that $x=y\times e^n$ for some integer $n$. Can you use that to find an integer $m$ such that $y=x\times e^m$? – lulu Feb 15 '23 at 23:56
  • i am unfamiliar with the notation ∼, does it just mean that (x,xe^n) => (xe^n,x), if so would the integer m = 0 prove it is symmetric. Many thanks by the way – jnljjkbj Feb 16 '23 at 00:20
  • Sorry. Given a relation $R$, writing $a\sim b$ or $a\sim_R b$ is the same as writing $(a,b)\in R$. so my writing $x\sim y$ means that $(x,y)\in R$ or, equivalently, that $y=x\times e^n$ for some integer $n$. Note that it is not true that $y=x\times e^n$ implies that $x=y\times e^n$. The two integers will not usually be the same (but they are closely related). – lulu Feb 16 '23 at 00:22
  • i have done more work on it, but am unsure if it is correct. If possible please may you provide help. Many thanks – jnljjkbj Feb 16 '23 at 00:46
  • No...again, for symmetry you'll need to think about the relation between $x$ and $y$. If $(x,y)\in R$ then we know that $y=x\times e^n$ for some integer $n$. But if that's true, then $x=y\times e^{-n}$ and, of course, $-n$ is still an integer. So we have $(y,x)\in R$ as desired. – lulu Feb 16 '23 at 01:13
  • You should not say $(x,x)\in R$ as that is what you need to prove. You should flip your argument around by changing so to because. Similarly for symmetric you need to prove that if $(x,y) \in R, (y,x) \in R$. For transitive the exponent witnessing $(x,y)$ and $(y,z)$ may be different, so you should use $n,m$ for them. Again the writing is backwards. – Ross Millikan Feb 16 '23 at 01:24

1 Answers1

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To summarize the discussion in the comments:

Reflexivity: $(x,x)\in R$ for all $x\in \mathbb R$ since $x=x\times e^0$.

Symmetry: $(x,y)\in R\implies y=x\times e^n$ for some integer $n$. But then $x=y\times e^{-n}$ and, of course, $-n$ is still an integer. Thus $(y,x)\in R$ as desired.

Transitivity: $(x,y)\,\&\,(y,z)\in R$ means that there are integers $n,m$ with $y=x\times e^n$ and $z=y\times e^m$. But in that case we have $z=x\times e^n\times e^m=x\times e^{n+m}$. And of course $n+m$ is still an integer, so we have $(x,z)\in R$.

And we are done.

lulu
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